Electronic Journal of Qualitative Theory of Differential Equations 2006, No.4, 1-19;http://www.math.u-szeged.hu/ejqtde/
A class of Second Order BVPs On Infinite Intervals
Sma¨ıl DJEBALI and Toufik MOUSSAOUI
Abstract
In this work, we are concerned with a boundary value problem asso- ciated with a generalized Fisher-like equation. This equation involves an eigenvalue and a parameter which may be viewed as a wave speed.
According to the behavior of the nonlinear source term, existence re- sults of bounded solutions, positive solutions, classical as well as weak solutions are provided. We mainly use fixed point arguments.
1 INTRODUCTION
The aim of this paper is to prove existence theorems for the boundary value problem
( −u00+cu0+λu = h(x, u), −∞< x <+∞.
|x|→lim+∞u(x) = 0. (1.1)
The parameter c > 0 is a real positive constant while h:R×R → R is a continuous function satisfying lim
|x|→+∞h(x,0) = 0; the parameter λ >0 may be seen as an eigenvalue of the problem. In the linear caseh(x, u) =f(x)u, the problem arises in the study of a reaction-diffusion system involved in disease propagation throughout a given population [2]; the sublinear case h(x, u) ≤ f(x)u was also studied in [2] where existence and non existence results are given. For other recent developments in solvability to boundary value problems on unbounded domains see [1, 7] and references therein.
In this work, we investigate the nonlinear case; the study of problem (1.1) depends on the growth type of the source term h with respect to the second argument. In Section 2, we prove existence of bounded classical solu- tions in case the nonlinear right-hand termhobeys a generalized polynomial growth condition. Section 3 is devoted to proving existence of positive so- lutions under integral restrictions on the nonlinear function h. A general
2000 Mathematics Subject Classification. 34B05, 34C20.
Key words. BVPs, Fixed point Theorems, Compactness Criterion
existence principle is given in Section 4. In Section 5, we show existence of positive solutions on the half-line under polynomial-like growth condition on the function h. Finally, existence of weak solutions is discussed in Section 6.
Our arguments will be based on fixed point theory. So, let us recall for the sake of completeness, respectively Schauder’s and Schauder-Tichonoff’s fixed point theorems [11]:
Theorem 1. Let E be a Banach space and K ⊂ E a bounded, closed and convex subset of E. Let F : K −→ K be a completely continuous operator. ThenF has a fixed point inK.
Theorem 2. Let K be a closed, convex subset of a locally convex, Hausdorff space E. Assume that T:K −→ K is continuous, and T(K) is relatively compact in E. ThenT has at least one fixed point inK.
In sequel, Ck(I,R) (k ∈ N) will refer to the space of kth continuously differentiable functions defined on an interval I of the real line. C0(R,R) stands for the space of continuous functions defined on the real line and vanishing at infinity; throughout this article, we will shorten the notation of this space to E0. Endowed with the sup-norm kuk= supx∈R|u(x)|, it is a Banach space. Recall thatLp(R) is the Banach space ofpthpower integrable functions onR. Hereafter, R+
∗ refers to the set of positive real numbers and the notation : = means throughout to be defined equal to.
2 A GENERALIZED POLYNOMIAL GROWTH CONDITION
The main existence result of this section is
Theorem 2.1 The Green function being defined by (2.3), assume the fol- lowing assumptions hold true:
∃Ψ : [0,+∞[−→[0,+∞[ continuous and nondecreasing;
∃q ∈E0 positive, continuous such that
|h(x, u)| ≤q(x)Ψ(|u|), ∀(x, u)∈R2;
∃M0 ∈R+
∗, αΨ(MM 0)
0 ≤1
withα: = supx∈R
R+∞
−∞ G(x, y)q(y)dy <∞.
(2.1)
Then Problem (1.1) admits a solution u∈E0.
Example 2.1 Consider a separated-variable nonlinear function h(x, y) = f(x)g(y) with |f(x)| ≤ x21+1: = q(x) and |g(y)| ≤ p
|y|+ 1: = Ψ(|y|). The real numbersr1, r2 being defined in (2.4), we have0≤G(x, y)≤ r1−1r2; then the constantα introduced in Assumptions 2.1 satisfies the estimate 0< α≤
1 r1−r2
R+∞
−∞ 1
x2+1dx= r π
1−r2.We infer the existence of some positive number M0 large enough such that r π
1−r2
√M0+ 1 ≤ M0. Therefore, αΨ(M0) ≤
π
r1−r2Ψ(M0) ≤M0 and so Assumptions 2.1 are satisfied. For instance, the following problem has at least one nontrivial solution:
( −u00+cu0+λu = (x2+1)(u1 2+1), −∞< x <+∞.
|x|→lim+∞u(x) = 0.
Remark 2.1 (a)Assumptions (2.1) encompass the case where the nonlin- ear function h satisfies the polynomial growth condition
∃f ∈E0, ∃ρ >0, |h(x, y)| ≤ |f(x)||y|ρ, ∀(x, y)∈R2
with either (ρ6= 1) or (ρ= 1 and supx∈R|f(x)| ≤λ). (2.2) (b) If, in Assumptions (2.1), the function h rather satisfies |h(x, y)| ≤
|f(x)| |y|ρ+β, ∀(x, y)∈R2, with some (f, β)∈E02, then we can only take ρ <1in Assumption (2.2). This particular case was studied in [9]; Theorem 2.1 then improves a similar result obtained in [9].
(c) Since we work on the whole real line, Theorem 2.1, as well as the other existence theorems in this paper, provide solutions which are not in general known to be nontrivial. To ensure existence of nontrivial solutions, one must add assumptions on the nonlinear function h such as h(x,0) 6≡0 further to lim
|x|→±∞h(x,0) = 0. Example 2.1 shows existence of at least one nontrivial solution.
(d) If we consider instead the autonomous case h(x, u) = g(u) with g(0) = 0, then the trivial solution u≡0is the unique solution. Let us prove this in two steps:
• For any solution u to Problem (1.1), note that lim
|x|→+∞u0(x) = 0. We check this whenx→+∞. Indeed, let
`: = lim inf
x→+∞u0(x)≤`: = lim sup¯
x→+∞
u0(x).
Then by a classical fluctuation lemma [6], there exist two sequences(xn)n∈N
and (yn)n∈N converging to positive infinity such that ` = lim
n→∞u0(xn) and
`¯= lim
n→∞u0(yn) whereas lim
n→∞u00(xn) = lim
n→∞u00(yn) = 0. Inserting into the equation in Problem (1.1), we find that cl = c¯l = 0; hence ` = ¯` = 0 for c >0.
• Define the energy E(x) = 1
2|u0(x)|2−λ
2|u(x)|2+G(u(x)) withG(u): =Ru
0 g(s)ds.Then, multiplying the equation in Problem (1.1) by u0 and making an integration by part, we find that E(x) =cRx
−∞|u0|2dxand so E0(x) =c|u0(x)|2 ≥0. As lim
x±∞E(x) = 0, we deduce that E is identically zero. From u0(x)2 = 1cE0(x) = 0, u is constant and hence u≡0.
(d) As for the separated-variable case h(x, u) =f(x)g(u), we must im- poseg(0)6= 0 both with f(±∞) = 0 otherwise we could also obtain u≡0 as a solution.
Under Hypothesis (2.1), we will make use of Schauder’s fixed point theorem to prove existence of a solution in a closed ball B(0, R) with some radius R >0.
Proof of Theorem 2.1 It is clear that Problem (1.1) is equivalent to the integral equation:
u(x) =R+∞
−∞ G(x, y)h(y, u(y))dy with Green function
G(x, y) = 1 r1−r2
er1(x−y) if x≤y
er2(x−y) if x≥y (2.3) and characteristic roots
r1 = c+√
c2+ 4λ
2 and r2 = c−√
c2+ 4λ
2 · (2.4)
Define the mappingT:E0→ E0 by T u(x) =
Z +∞
−∞
G(x, y)h(y, u(y))dy. (2.5) In view of Schauder’s fixed point theorem, we look for fixed points for the operator T in the Banach spaceE0. The proof is split into four steps.
• Claim 1: The mappingT is well defined; indeed, for any u∈E0, we get, by Assumptions (2.1), the following estimates:
|T u(x)| ≤ R+∞
−∞ G(x, y)|h(y, u(y))|dy
≤ R+∞
−∞ G(x, y)q(y)Ψ(|u(y)|)dy
≤ Ψ(kuk)R+∞
−∞ G(x, y)q(y)dy, ∀x∈R
≤ αΨ(kuk).
The convergence of the integral definingT u(x) is then established. In addi- tion for any y ∈ R, G(±∞, y) = 0, and then, taking the limit in T u(x), we get, by l’Hospital Theorem, T u(±∞) = 0. Therefore, the mapping T :E0 →E0 is well defined.
Claim 2: The operator T is continuous.
Let be a sequence (un)n∈E0converge uniformly tou0on all compact subin- terval of R. For some fixed a > 0, we will prove the uniform convergence of (T un)n to some limit T u0 on the interval [−a, a]. Let ε > 0 and choose someb > alarge enough. By the uniform convergence of the sequence (un)n on [−b, b], there exists an integer N =N(ε, b) satisfying
n≥N =⇒I1: = sup
x∈R
Z +b
−b
G(x, y)|h(y, un(y))−h(y, u0(y))|dy < ε 2· Forx∈[−a, a],we have that|T un(x)−T u0(x)| ≤I1+I2+I3 with:
I2: = supx∈R
R
R−[−b,+b]G(x, y)|h(y, u0(y))|dy ≤ ε4 (by Cauchy Convergence Criterion and lim
|y|→+∞h(y, u0(y)) = 0.) I3: = supx∈R
R
R−[−b,+b]G(x, y)|h(y, un(y))|dy ≤ ε4· (by Lebesgue Dominated Convergence Theorem.)
This proves the uniform convergence of the sequence (T un)n to the limit T u0 on the interval [−a, a].
Claim 3: For any M > 0, the set {T u,kuk ≤ M} is relatively com- pact in E0. By Ascoli-Arzela Theorem, it is sufficient to prove that all the functions of this set are equicontinuous on every subinterval [−a, a] and that there exists a functionγ ∈E0 such that for anyx∈R,|T u(x)| ≤γ(x). Let x1, x2∈[−a, a] ; we have successively the estimates:
|T u(x2)−T u(x1)| ≤ R+∞
−∞ |G(x2, y)−G(x1, y)||h(y, u(y))|dy
≤ R+∞
−∞ |G(x2, y)−G(x1, y)|q(y)Ψ(|u(y)|)dy
≤ Ψ(M)R+∞
−∞ |G(x2, y)−G(x1, y)|q(y)dy.
By continuity of the Green functionG, the latter term tends to 0, whenx2 tends x1; whence comes the equicontinuity of the functions {T(u);kuk ≤ M}. Now, we check analogously the second statement:
|T u(x)| ≤ R+∞
−∞ G(x, y)|h(y, u(y))|dy
≤ R+∞
−∞ G(x, y)q(y)Ψ(|u(y)|)dy
≤ Ψ(M)R+∞
−∞ G(x, y)q(y)dy: =γ(x), ∀x∈R. By l’Hopital Theorem, we have thatγ ∈E0.
Claim 4: There exists some R > 0 such that T maps the closed ball B(0, R) into itself. From assumption (2.1), we know that there is some positive numberM0 such that αΨ(MM 0)
0 ≤1. If kuk ≤M0, then kT(u)k ≤ supx∈R
R+∞
−∞ G(x, y)q(y)Ψ(|u(y)|)dy
≤ αΨ(M0)
≤ M0,
so that it is enough to takeR=M0. The proof of Theorem 2.1 then follows from Schauder’s fixed point theorem.
3 EXISTENCE OF POSITIVE SOLUTIONS
Making use of Schauder-Tichonov’s theorem, we prove here existence of a positive solution under an integral condition on the nonlinear term:
Theorem 3.1 Problem (1.1) has a positive solution provided the following mean growth assumption on the nonlinear function h is fulfilled
The function h is positive and satisfiesh(x, u)≤H(x,|u|) whereH: R×R+ → R+is continuous, nondecreasing
with respect to the second argument and verifies
∃c∗ >0, R+∞
−∞ H(x, c∗)dx≤c∗(r1−r2).
(3.1)
Remark 3.1 It is easy to check that in the separated-variable case, As- sumption (3.1) leads to Assumption (2.1).
Example 3.1 The problem
( −u00+cu0+λu = x2u+un 2 +x21+1, (n∈N, n >2) − ∞< x <+∞;
|x|→lim+∞u(x) = 0
has at least one positive nontrivial solution. Indeed, the function H(x, y) =
yn
x2+y2 +x21+1 satisfies lim
|x|→0H(x,0) = 0,H(x,0)6≡0 and is nondecreasing in the second argument y for any integer n >2. Moreover, R+∞
−∞ H(x, y)dx= π(1 +yn−1) so that Assumption (3.1) may be satisfied. For instance, if we take n= 3, then there exists c∗ ∈]c1, c2[ with c1 = k−√2k2−1, c2 = k+√2k2−1 assumingk: = r1−πr2 >2.
To prove Theorem 3.1, we proceed as in Theorem 2.1, and reformulate Prob- lem (1.1) as a fixed point problem for the mappingT defined in 2.5. Here, we appeal to Schauder-Tichonov’s fixed point theorem. LetK be the closed convex subset ofE0 defined by:
K ={u∈E0, 0≤u(x)≤c∗, ∀x∈R}.
Using Assumption (3.1) and the fact that the mappingH is nondecreasing in the second argument, we find thatT maps K into itself. Indeed, taking into account the bound 0< G(x, y) ≤ r1−1r2, we derive the straightforward estimates:
0≤T u(x) ≤ R+∞
−∞ G(x, y)H(y,|u(y)|)dy
≤ r1−1r2
R+∞
−∞ H(y, c∗)dy
≤ c∗. Since 0≤T u(x) ≤R+∞
−∞ G(x, y)H(y, c∗)dy and G(±∞, y) = 0, ∀y∈R,we have that T u(±∞) = 0 and so T(E0)⊂E0.In addition, the mapping T is continuous as can easily be seen. It remains to check thatT(K) is relatively compact. By Ascoli-Arzela Theorem, it is sufficient to prove that all the functions of this set are equicontinuous on every subinterval [−a, a] and that there exists a functionγ ∈E0 such that for anyx∈R,|T u(x)| ≤γ(x).
Letx1, x2 ∈[−a, a] ;
|T u(x2)−T u(x1)| ≤ R+∞
−∞ |G(x2, y)−G(x1, y)||h(y, u(y))|dy
≤ R+∞
−∞ |G(x2, y)−G(x1, y)|H(y, c∗)dy.
By continuity of the function G, we deduce from Lebesgue dominated con- vergence theorem that the last right-hand term tends to 0 whenx2 tends to x1. Whence comes the compactness of T(K) by Ascoli-Arzela Lemma and then the claim of Theorem 3.1 follows.
Now, we check analogously the second statement:
|T u(x)| ≤ Z +∞
−∞
G(x, y)H(y, c∗)dy ≡γ(x) withγ ∈E0 forG(±∞, y) = 0, ∀y∈R.
4 A FURTHER TYPE OF GROWTH
In this section, we prove existence of bounded, solutions to Problem (1.1) under new growth conditions on the nonlinearity h; by the way we show that polynomial-like growth condition may be relaxed. The proof of our existence result relies on the following fixed point theorem by Furi and Pera [3]. This theorem was also used in [1] to deal with a BVP on an infinite interval.
Theorem 3. Let E be a Fr´echet space, Q a closed convex subset of E, 0∈ Q and let T :Q → E be a continuous compact mapping. Assume further that, for any sequence (uj, µj)j≥1 from ∂Q×[0,1] that converges to (u, µ) withu=µT u, 0≤µ < 1,one hasµjT uj ∈Q for all j large enough.
Then,T has a fixed point in Q.
Our aim is now to prove
Theorem 4.1 First, assume the following sign assumption is fulfilled:
(H1) ∃M0 >0 such that |y|> M0 ⇒ yh(x, y) ≤0, ∀x∈R. Then Problem (1.1) has a bounded solution provided either one of the following growth assumptions is satisfied:
(H2)1 There exists a functionH : R×R+ → R+ continuous and nondecreasing with respect to the second argument such that:
|h(x, y)| ≤H(x,|y|), ∀(x, y)∈R2 with R+∞
−∞ H(x, M0+ 1)dx <∞; or (H2)2 |h(x, y)| ≤q(x)ψ(|y|), ∀(x, y)∈R2
withψ: R+ → R+ continuous and nondecreasing, q∈E0 positive, continuous and α: = supx∈R
R+∞
−∞ G(x, y)q(y)dy <∞; or (H2)3 lim
x→+∞sup|y|≤M0+1|h(x, y)|= 0.
Example 4.1 Consider the function h defined by
h(x, y) =
1−y
x2+1, y≥+1
1
x2+1, −1≤y≤+1
−1−y
x2+1, y≤ −1
Then,|h(x, y)| ≤ 1+x2+1|y|=H(x,|y|)withH(x, s) = x1+s2+1·SinceR+∞
−∞
2
x2+1dx= 2π < ∞, Assumptions (H1) and (H2)1 are satisfied with M0 = 1. Since h(x,0) 6≡ 0, Problem (1.1) has a nontrivial solution for such a nonlin- ear right-hand term. We may notice that the function h can be written ash(x, y) = θ(y)x2+1−y whereθ(y) =H(y−1)−H(−y−1), the function Hbeing the Heaviside function.
Remark 4.1 (a) The sign condition (H1) implies that any solution u of Problem (1.1) satisfies |u(x)| ≤ M0, ∀x ∈ R. Indeed, on the contrary, assume maxx∈R|u(x)|= |u(x0)| > M0 for some x0 ∈ R. Then u0(x0) = 0 and −u00(x0)u(x0) +cu0(x0)u(x0) +λu(x0)2 = u(x0)f(x0, u(x0)) ≤ 0; this yieldsu00(x0)u(x0)≥0, leading to a contradiction.
(b) Assumption (H2)1 is weaker than the one in Theorem 3.1.
(c) Assumption (H2)2 is weaker than the one in Theorem 2.1.
(d) Then, when Assumption (H1) is satisfied, Theorem 4.1 improves both of Theorems 2.1 and 3.1.
Proof of Theorem 4.1. For the sake of clarity, we only do the proof in case Assumptions (H1) and (H2) are simultaneously satisfied. The other cases can be treated similarly. In the Fr´echet spaceE: =C(R,R), setr0: =M0+1, consider the closed, convex set Q={u ∈E : supx∈R|u(x)| ≤r0}, and de- fine the mapping T:Q → E as in (2.5). In three steps, we carry over the
proof.
• Claim 1. T is continuous. Let (un)n∈N be a sequence in Q such that un → u in Q; we show that T un → T u in Q, as n goes to infinity. We have that for any x ∈ R, |h(x, un(x))| ≤ H(x, r0), |h(x, u(x))| ≤ H(x, r0) and that lim
n→∞h(x, un(x)) = h(x, u(x)). By the Dominated Convergence Lebesgue’s Theorem, we deduce that lim
n→∞T un(x) =T u(x) on each subin- terval [−xm, xm].
In addition, for x1, x2∈[−xm, xm], the following estimates hold true
|T un(x2)−T un(x1)| ≤ R+∞
−∞ |G(x2, y)−G(x1, y)|H(y, r0)dy;
|T u(x2)−T u(x1)| ≤ R+∞
−∞ |G(x2, y)−G(x1, y)|H(y, r0)dy.
Therefore,∀ε >0, ∃δ >0 such that
|x2−x1|< δ ⇒ |T u(x2)−T u(x1)|< ε and|T un(x2)−T un(x1)|< ε,∀n∈N. Furthermore, the convergence is uniform sinceT un(x) → T u(x) on [−xm, xm] asn→+∞, and the claim follows.
• Claim 2. Using Ascoli-Arzela Theorem, we are going to prove that T(Q) is relatively compact in E, that is T(Q) is uniformly bounded and equicontinuous on each subinterval [−xm, xm]. For any x∈ [−xm, xm] and any u∈Q, we have:
|T u(x)| ≤ R+∞
−∞ G(x, y)H(y,|u(y)|)dy
≤ R+∞
−∞ G(x, y)H(y, r0)dy: =ψr0(x), with, by l’Hospital rule, lim
x→±∞ψr0(x) → 0; that is ψr0 ∈ E0. Moreover, T(Q) is equicontinuous on each subinterval [−xm, xm]. Indeed, let x1, x2 ∈ [−xm, xm] withx1< x2; we have:
|T u(x2)−T u(x1)| ≤ Z +∞
−∞ |G(x2, y)−G(x1, y)|H(y, r0)dy which also tends to 0 asx2 tends tox1, for anyu∈Q.
•Claim 3. Now, we check the last assumption in Furi-Pera’s Theorem.
Consider some sequence (uj, µj)j≥0 ∈ ∂Q×[0,1] such that, when j → ∞, µj → µ and uj → u with u = µT u and µ ∈ [0,1]. We must show that µjT uj ∈Qasj→ ∞. Letv∈E be such that|v(x)| ≤r0forx∈R, then we have that |T v(x)| ≤R+∞
−∞ G(x, y)H(y, r0)dy: =ψr0(x) and lim
|x|→+∞ψr0(x) = 0. Sinceuj ∈Q, there exists somex∗∈Rsuch that
∀x∈R\[−x∗, x∗], |µjT uj(x)| ≤r0.
Let us consider the casex∈[−x∗, x∗]. Sinceµj →µandT(Q) is bounded in E, the sequenceµjT uj converges uniformly toµT uon [−x∗, x∗]; thus there exists some j0 ∈ N∗ such that ∀j ≥ j0, |µjT uj(x)| ≤ |µT u(x)|+ 1, ∀x ∈ [−x∗, x∗]. We have also by Remak 4.1(a) that |µT u(x)| ≤ M0. Therefore, forj large enough, it holds that
|µjT uj(x)| ≤M0+ 1 =r0, ∀x∈[−x∗, x∗].
We then conclude the estimate|µjT uj(x)| ≤M0+ 1 =r0, ∀x∈R. The claim of Theorem 4.1 then follows from Theorem 3.
5 THE PROBLEM ON THE POSITIVE HALF LINE
5.1 Setting of the problem
In this section, we consider the problem posed on the positive half-line:
−u00+cu0+λu = h(x, u(x)), x∈I
u(0) =u(+∞) = 0. (5.1)
Hereafter, I denotes ]0,+∞[, the set of positive real numbers. Setting k: =p
λ+c2/4,we rewrite the problem for the function v(x) =e−2cxu(x):
(
−v00+k2v = e−c2 xh(x, ec2xv(x)), x∈I
v(0) =v(+∞) = 0. (5.2)
Equivalently, the unknownv satisfies the integral equation:
v(x) =R+∞
0 K(x, s)e−2csh(s, ec2sv(s))ds with new Green function
K(x, s) = 1 2k
e−ks(ekx−e−kx) x≤s
e−kx(eks−e−ks) x≥s. (5.3) Notice that K is different from the one in (2.3) and that the unknownu is now solution of the integral equation:
u(x) =R+∞
0 ec2(x−s)K(x, s)h(s, u(s))ds.
The following lemma provides estimates of the Green function K and will play an important role in the sequel; we omit the proof:
Lemma 5.1 We have
(a) K(x, s)≤ 2k1 , K(x, s)e−µx≤K(s, s)e−ks, ∀x, s∈I, ∀µ≥k.
(b) ∀s∈I,∀(0< γ < δ), ∀x∈[γ, δ], K(x, s)≥mK(s, s)e−ks. (5.4) Here m: = min
e−kδ, ekγ−e−kγ .
Under suitable assumptions on the nonlinear function h, we shall prove the existence of a positive solution to Problem (5.1). The proof relies on Krasnosels’kii fixed point theorem in cones ([5], [8]) and Zima’s compactness criterion [12]; but first of all, let us recall some
5.2 Preliminaries
Definition 5.1 A nonempty subset C of a Banach space X is called a cone if C is convex, closed, and satisfies
(i) αx∈ C for all x∈ C and any real positive numberα, (ii) x,−x∈ C imply x= 0.
Definition 5.2 A set of functionsu∈Ω⊂Xare said to be almost equicon- tinuous onI if they are equicontinuous on each interval[0, T], 0< T <+∞. Next we state Krasnosels’kii Fixed Point Theorem in cones.
Theorem 4. ([8]) LetX be a Banach space andC ⊂X be a cone inX.
Assume that Ω1 and Ω2 are two bounded open sets in X such that 0 ∈Ω1 and ¯Ω1 ⊂ Ω2. Let F : C ∩(Ω2 −Ω1) −→ C be a completely continuous operator such that either
(i) kF xk ≤ kxk forx∈ C ∩∂Ω1 and kF xk ≥ kxk forx∈ C ∩∂Ω2, or (ii) kF xk ≥ kxkforx∈ C ∩∂Ω1 and kF xk ≤ kxkforx∈ C ∩∂Ω2 is satisfied.
Then F has at least one fixed point in C ∩(Ω2−Ω1).
Now, let p:I −→]0,+∞[ be a continuous function. Denote by X the Banach space consisting of all weighted functions u continuous on I and satisfying
supx∈I{|u(x)|p(x)}<∞,
equipped with the normkuk= supx∈I{|u(x)|p(x)}.We have
Lemma 5.2 ([13]) If the functions u ∈Ω are almost equicontinuous on I and uniformly bounded in the sense of the norm
kukq= supx∈I{|u(x)|q(x)}
where the functionq is positive, continuous on I and satisfies
x→lim+∞ p(x) q(x) = 0, then Ω is relatively compact inX.
Having disposed of these auxiliary results, we are ready to prove
Theorem 5.1 Suppose that:
h:I×R+−→R+ is a continuous function,
∃p >0 : p6= 1, h(x, u)≤a(x) +b(x)up, ∀(x, u)∈I×R+, where a, b:I −→R+ are continuous positive functions vanishing at positive infinity and
(5.5)
there exists θ > k+2c such that M1: =R+∞
0 e−(k+c2)sa(s)ds <∞, M2: =R+∞
0 e(pθ−k−c2)sb(s)ds <∞.
(5.6)
2k 2k
pM2 p−11
−M2
2k pM2
p−1p
−M1 ≥0, when p >1. (5.7)
2k 2k
pM2 p−11
−M2 2k
pM2 p−1p
−M1 ≤0, when 0< p <1. (5.8) ( There exist α >0, γ, δ >0 and x0 ∈I such that:
minx∈[γ,δ],u∈[mα,αeθδ]h(x, u)≥αeθx0h Rδ
γ ec2(x0−s)K(x0, s) dsi−1
, (5.9) the Green functionKbeing defined in (5.3) andm: = min
e−kδ, ekγ−e−kγ . Then Problem (5.1) has at least one positive solution u∈C(I;R+).
Remark 5.1 The case p= 1 is treated in [2].
Proof of Theorem 5.1:
We follow the method used in [2, 14]. Letθ∈Rbe as in Hypothesis (5.6) and consider the weighted space X = {u ∈ C(I;R): supx∈I{e−θx|u(x)|} <∞}
endowed with the weighted sup-norm:
kukθ= supx∈I{e−θx|u(x)|}, as well as the positive cone inX:
C={u∈X;u≥0 on I and minx∈[γ,δ]u(x)≥mkukθ}. Next define onC the operatorF by:
F u(x) =R+∞
0 ec2(x−s)K(x, s)h(s, u(s))ds.
(a) First step. In the following, we study the properties of this operator:
• Claim 1:
For anyu∈X, supx∈Ie−θx|F u(x)|<∞,that is F(X)⊂X.
Indeed, choosingµ=θ−2c in (5.4) (note that µ≥kby (5.6)), and using (5.4)(a), (5.5), we have the estimates
0≤e−θxF u(x) =
=e−θxR+∞
0 ec2(x−s)K(x, s)h(s, u(s))ds
=e−(θ−c2)xR+∞
0 e−c2 sK(x, s)h(s, u(s))ds
≤R+∞
0 e−(k+2c)sK(s, s) [a(s) +b(s)|u(s)|p]ds
≤ 2k1 R+∞
0 e−(k+c2)sa(s)ds + 2k1 kukθ
R+∞
0 e(pθ−k−c2)sb(s)ds
≤ 2k1 (M1+M2kukpθ)<∞.
• Claim 2:
For anyu∈ C, let us check that minx∈[γ,δ]F u(x)≥mkF ukθ that isF(C)⊂ C. Indeed,F u(x)≥0 ∀x∈I and ∀x∈[γ, δ],∀s, τ ∈I we have, choosing µ=θ−2c in (5.4)(a):
minx∈[γ,δ]F u(x) ≥ mR+∞
0 ec2(γ−s)K(s, s)e−ksh(s, u(s))ds
≥ meγc2e−θτR+∞
0 ec2(τ−s)K(τ, s)h(s, u(s))ds
≥ mec2γkF ukθ ≥mkF ukθ. Next, we prove thatF is completely continuous:
• Claim 3:
Let Ω1 = {u ∈ X,kukθ < r},Ω2 = {u ∈ X,kukθ < R}, the constants 0 < r < R being real positive numbers to be selected later on. Consider someu∈ C ∩Ω2; then F u is uniformly bounded. Indeed, as in claim 1, we have thatkF ukθ≤ 2k1(M1+M2Rp),for anyu∈ C ∩Ω2.
• Claim 4:
The functions{F u}foru∈ CK∩Ω2are almost equicontinuous onI; indeed
|F u(x2)−F u(x1)| ≤
Z +∞
0
e2c(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds
= Z x1
0
ec2(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds +
Z x2 x1
ec2(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds +
Z +∞ x2
e2c(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds.
In the following, we derive lengthy estimates of each of the summands in the right-hand side. We have successively:
Z x1 0
ec2(x2−s)K(x2, s)−e2c(x1−s)K(x1, s)
h(s, u(s))ds≤
≤ 1 2k
e(c2−k)x2 −e(c2−k)x1
Z x1
0
e−c2s(eks−e−ks)a(s)ds
+ 1
2k
e(c2−k)x2 −e(c2−k)x1
Z x1 0
e−c2s(eks−e−ks)b(s)|u(s)|pds
≤ 1 2k
e(c2−k)x2 −e(c2−k)x1
Z x1
0
e−c2s(eks−e−ks)a(s)ds
+ 1
2k
e(c2−k)x2 −e(c2−k)x1 kukpθ
Z x1 0
e(pθ−c)s(eks−e−ks)b(s)ds
≤ 1 2k
e(c2−k)x2 −e(c2−k)x1
Z x1
0
e−c2s(eks−e−ks)a(s)ds
+ 1
2kRp
e(c2−k)x2 −e(2c−k)x1
Z x1
0
e(pθ−c2)s(eks−e−ks)b(s)ds.
The right-hand term in the last inequality tends to 0 when x2 −→ x1, for any u∈ C ∩Ω¯2. In addition, we have the bounds:
Z x2 x1
ec2(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds≤
≤ 1
2ke(2c−k)x2 Z x2
x1
e−c2s(eks−e−ks)a(s)ds
+ 1
2ke(2c−k)x2kukpθ
Z x2
x1
e(pθ−c2)s(eks−e−ks)b(s)ds
+ 1
2kec2x1(ekx1−e−kx1) Z x2
x1
e−(c2+k)sa(s)ds
+ 1
2kec2x1(ekx1−e−kx1)kukpθ
Z x2
x1
e(pθ−2c−k)sb(s)ds
≤ 1
2ke(2c−k)x2 Z x2
x1
e−c2s(eks−e−ks)a(s)ds
+ 1
2ke(2c−k)x2Rp Z x2
x1
e(pθ−c2)s(eks−e−ks)b(s)ds
+ 1
2kec2x1(ekx1−e−kx1) Z x1
x2
e−(c2+k)sa(s)ds
+ 1
2kec2x1(ekx1−e−kx1)Rp Z x2
x1
e(pθ−2c−k)sb(s)ds.
Again, all of the terms in the right side tend to 0 when x2 −→ x1, for all u∈ C ∩Ω¯2.
At last, we have:
Z +∞
x2
ec2(x2−s)K(x2, s)−ec2(x1−s)K(x1, s)
h(s, u(s))ds≤
≤ 1 2k
ec2x2(ekx2 −e−kx2)−ec2x1(ekx1 −e−kx1)
Z +∞
x2
e−(c2+k)sa(s)ds
+ 1
2k
ec2x2(ekx2 −e−kx2)−ec2x1(ekx1 −e−kx1) kukpθ
Z +∞
x2
e(pθ−c2−k)sb(s)ds
≤ 1 2k
ec2x2(ekx2 −e−kx2)−ec2x1(ekx1 −e−kx1)
Z +∞ x2
e−(c2+k)sa(s)ds
+ 1
2k
ec2x2(ekx2 −e−kx2)−ec2x1(ekx1 −e−kx1) Rp
Z +∞ x2
e(pθ−c2−k)sb(s)ds.
And all of the terms in the right side tend to 0 when x2 −→ x1, for all u∈ C ∩Ω¯2.
According to Lemma 5.2, we conclude that the operatorF is completely continuous onC ∩Ω2.
(b) Second step. Now, we check the first alternative in Theorem 4.
•Ifu∈ C∩∂Ω1, thene−θxF u(x)≤ 2k1 (M1+M2kukpθ)≤ 2k1(M1+M2rp)≤ r which is fulfilled by Assumptions (5.7), (5.8). We have then proved that kF ukθ ≤ kukθ.
• Moreover, if u ∈ C ∩∂Ω2, then take kukθ = R = α where α is as defined in Assumption (5.9) and find that minx∈[γ,δ]u(x)≥mα. Hence, for any x∈[γ, δ], mα≤u(x)≤αeθδ. Furthermore, it holds that:
F u(x0) = R+∞
0 ec2(x0−s)K(x0, s)h(s, u(s))ds
≥ Rδ
γ ec2(x0−s)K(x0, s)h(s, u(s))ds
≥ h
minx∈[γ,δ],u∈[mα,αeθδ]h(x, u)i
αeθx0Rδ
γ ec2(x0−s)K(x0, s) ds
≥ αeθx0.
Consequently,e−θx0F u(x0)≥α that iskF ukθ≥ kukθ for anyu∈K∩∂Ω2. Thanks to Theorem 4, the operatorF has a fixed point inC ∩(Ω2\Ω1) and so Problem (5.1) admits a positive solutionu in the cone C.
6 WEAK SOLUTIONS
In contrast with the previous results, we look here for solutions in the Lebesgue Space Lp(R). We first need a compactness criterion in Lp(R) due to Fr´echet-Kolmogorov:
Theorem 5. ([10], p. 275) A set S⊂Lp(R) (1≤p <+∞) is relatively compact if and only ifS is bounded and for every ε >0, we have:
(i)∃δ >0 such that R+∞
−∞ |u(x+h)−u(x)|pdx < ε, ∀u∈S, ∀0< h < δ.
(ii) There exists a numberN >0 such that R
R\[−N,N]|u(x)|pdx < ε, ∀u∈S.
Let us recall a
Definition 6.1 We say that f:I×R −→ R is a Carath´eodory function if (i) the map x −→ f(x, y) is measurable for all y∈R.
(ii) the map y −→ f(x, y) is continuous for almost everyx∈I.
(iii) there exists h∈L1(I) such that |f(x, y)| ≤h(x), for a.e. x∈I and for ally∈R.
Now, we are now in position to state an existence result for weak solu- tions:
Theorem 6.1 Assume the separated-variable nonlinear function h(x, u) = q(x)g(u) is of Carath´eodory type with q ∈ Lp(R) (1 < p < +∞), and g satisfies the general polynomial growth condition:
∃k, σ >0, |g(y)| ≤k|y|σ, for a.e. x∈Rand for all y∈R. α: =R+∞
−∞
R+∞
−∞ |G(x, y)|pqp(y)dypσ−1
dx <∞,
with(θ6= 1) or (θ= 1 andkαp−1pσ ≤1) with θ: = (pp−21)σ2·
(6.1)
Then problem (1.1) has a solution in Lr(R) withr= ppσ−1·
Remark 6.1 The sublinear case σ = 1 was studied in [2] with p = 2; the solutions are then found to be L2(R).
Proof :
Consider the Banach spaceE =Lr(R) endowed with the usual normkukr = (R+∞
−∞ |u(s)|rds)1r; hereafter, the notationkukr will be shorten to kuk. The mapping T:E −→ E is as defined in subsection 2.1. We will make use of H¨older inequality kf gk1 ≤ kfkp.kgkp∗ with p∗ = p−p1 the conjugate of p, that is p∗ = rσ·
Claim 1: T is continuous:
Consider some u0 ∈Lr(R) and prove the continuity of T at u0. By H¨older inequality, we have:
|T u(x)−T u0(x)|r=
Z +∞
−∞
G(x, y)q(y)[g(u(y))−g(u0(y))]dy
r
≤
Z +∞
−∞ |G(x, y)|pqp(y)dy rp
.
Z +∞
−∞ |g(u(y))−g(u0(y))|p∗dy p∗r
.
